A) \[E=\frac{3}{4}\,\left( \frac{m{{v}^{2}}}{qa} \right)\]
B) The rate of work done by the electric field at P is \[\frac{3}{4}\left( \frac{m{{v}^{3}}}{a} \right)\]
C) The rate of work done by the electric field at P is 0.
D) The rate of work done by both the fields at Q is 0.
Correct Answer: C
Solution :
In going from P to Q increase in kinetic energy \[=\frac{1}{2}\,m{{(2v)}^{2}}-\frac{1}{2}\,m{{v}^{2}}=\frac{3}{2}\,(m{{v}^{2}})\] = work done by electric field or \[\frac{3}{2}\,m{{v}^{2}}=Eq\times 2a\] or \[E=\frac{3}{4}\,\left( \frac{m{{v}^{2}}}{qa} \right)\] The rate of work done by E at P = force due to E \[\times \] velocity \[=(qE)v=qv\,\left[ \frac{3}{4}\,\left( \frac{m{{v}^{2}}}{qa} \right) \right]=\frac{3}{4}\left( \frac{m{{v}^{3}}}{a} \right)\] \[\left( \text{as},\,\,P=\frac{dw}{dE}=Fv\,\,\cos \,\theta \right)\] At q, v is perpendicular to E and B and no work is done by either field.
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